Question

Random samples of size n = 2 are drawn from a finite population that consists of the numbers 2, 4, 6, and 8. (a) Calculate th
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Answer #1

Answer:

Given that,

\rightarrow Random samples of size n = 2 are drawn from a finite population that consists of the numbers 2,4,6, and 8.

(a).

Calculate the mean and the standard deviation of this population:

\rightarrow We have population: 2, 4, 6, 8

Mean (\mu)=(2+4+6+8)/4

=20/4

\mu=5

Standard deviation(\sigma):

We know that,

σ? Σε – μ)? η –1

\sigma^2=\frac{ (2-5 )^{2}+(4-5)^{2}+(6-5 )^{2}+(8-5 )^{2}}{4-1}

\sigma^2=\frac{ 9+1+1+9}{3}

\sigma^2=\frac{ 20}{3}

\sigma^2=6.6667

\sigma=\sqrt{6.6667}

\sigma=2.5819

(b).

List the six possible random samples of size n = 2 that can be drawn from this population and calculate their means:

Samples are given by:

Samples Mean
(2,4) 3
(2,6) 4
(2,8) 5
(4,6) 5
(4,8) 6
(6,8) 7

(c).

Use the results of part (b) to construct the sampling distribution of the mean for random samples of size n =2 from the given population:

Sampling distribution of the mean is:

\mu _{\bar{x}}=\mu =5

(d).

Calculate the standard deviation of the sampling distribution obtained in part c) and verify the result by substituting n = 2, N = 4, and the value of σ obtained in part (a):

We know that,

\sigma _{\bar{x}}=\frac{\sigma }{\sqrt{n}}

=\frac{2.5819 }{\sqrt{5}}

=2.5819/2.2361

=1.1546

For n=2:

\sigma _{\bar{x}}=\frac{\sigma }{\sqrt{n}}\times \sqrt{\frac{N-n}{N-1}}

=\frac{2.5819 }{\sqrt{2}}\times \sqrt{\frac{4-2}{4-1}}

=\frac{2.5819 }{1.4142}\times \sqrt{0.6667}

=1.8257\times 0.8165

\sigma _{\bar{x}} =1.4907 (Approximately)

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